Exam 2 Cheat Sheet
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Transcript of Exam 2 Cheat Sheet
φ
F
Fx
Fy
N
W
Fs
P-x- h
N
Fs
Rsφs
Chapter 11 lb = 4.448 N1 slug = 14.59 kg1 inch = 2.54 cm1 ft = 0.3048 m
Chapter 2Cosine Law: C=√ A2+B2−2 ABcosγ
Sine Law: A
sin (a )= B
sin ( β )= C
sin (γ )
FR=√FRx2 +FRy
2 , tan−1(F Ry
F Rx
)
Coordinate direction angles:
cos (α )=Ax
A, cos ( β )=
Ay
A,cos (γ )=
A z
A
uA=Ax
Ai+
A y
Aj+
A z
Ak
uA=cos (α )i+cos (β) j+cos (γ ) k
A' : projection of 3 D vector onto xy planeφ : angle of vector out ¿ zaxisθ :angle of vector out ¿x axisA z=A cos (φ ) , A '=A sin (φ )Ax=A' cos (θ )=A sin (φ ) cos(θ)A y=A ' sin (θ )=A sin (φ )sin(θ)Force directed along a line:
F=F u=F( ( xB−x A ) i+ ( yB− y A ) j+( z B−z A ) k
√( xB−x A )2+( yB− y A )2+( zB−z A )2 )Remember notation: r BA=r A−r B
Dot product: A ∙ B=AB cos (θ )¿ Ax Bx+ A y By+ A z B z
Chapter 3∑ F x=∑ F y=¿∑ F z=0¿ Linear Elastic Springs:F=ks where k=stiffness/spring constant,l0=unloaded length, l=loaded length,s=displacement (l-l0). +s=elongation
-s=compressionChapter 4Moments: Mo=Fd (where d is ┴ distance from F to O)Moment of F about O : MO=r × FM O=rF sin (θ )Rotation along specified axis, a:
M a=ua∙(r× F)=|uax uay uaz
r x r y rz
Fx F y F z|
Chapter 5 (Free Body Diagrams)
Chapter 6 (Methods of Joints & Sections / Frames and Machines)
Chapter 8 (Friction)N: resultant normal forceF: resultant friction forceP: applied forceW: weight, mgIf x=0 : slippageIf ∑ M o=0 ,Wx=Ph Static frictional force , F s normal force , N :
F s=μs N ,φs=tan−1( F s
N )=tan−1(μs)
Two options:1. Slipping before tipping2. Tipping before slippingWhen P>F s , F sdrops ¿ Fk
θ
F x=F cos (θ ) , F y=F sin (θ )F x=F sin (φ ) ,F y=F cos (φ )